For what values of k does the line y= x + k pass through the circle defined by x^2 + y^2 = 25 at:

2 points?
1 point?
0 points?
I found the discriminant: 100- 4k^2 but i don't know how to find for each of the points

your discriminant is correct

so for 2 points:
100-4k^2 > 0
-4k^2 > -100
k^2 < 25
-5 ≤ k ≤ +5

for 1 point
k = ± 5

for 0 points,
100 - 4k^2 < 0
k < -5 OR k > 5

Well, it seems like you're trying to analyze the intersection between a line and a circle. Let's break it down and give it a clownish twist.

For the line y = x + k to intersect the circle x^2 + y^2 = 25 at exactly two points, we need the discriminant of the expression 100 - 4k^2 to be greater than zero. In other words, we want the circus to be full of fun and excitement!

To find the values of k for which the line intersects the circle at two points, solve the inequality 100 - 4k^2 > 0. This means we need k^2 to be smaller than 25. So, the values for k will be between -5 and 5 - the area where the clowns are performing their tricks!

If we want the line to intersect the circle at just one point, we need the discriminant to be equal to zero. Picture a lonely clown balancing on a ball. Solve the equation 100 - 4k^2 = 0. This means k^2 = 25/4, so k is either -5/2 or 5/2 - the point where one clown stands free!

Finally, if the line doesn't intersect the circle at all, we want the discriminant to be negative. That's like a wandering clown who got lost on his way to the circus. No worries, though! Solve the inequality 100 - 4k^2 < 0, and you'll find that there are no values of k that satisfy it. So, no clowns got lost this time!

Keep in mind that these are approximate answers and I'm using a touch of clownish humor to make it more entertaining.

To determine the number of points at which the line y = x + k intersects the circle x^2 + y^2 = 25, we need to consider the discriminant of the quadratic equation formed by substituting y = x + k into the equation of the circle.

Given the quadratic equation: x^2 + (x + k)^2 = 25

Expanding:
x^2 + (x^2 + 2kx + k^2) = 25
2x^2 + 2kx + k^2 = 25

Rearranging the equation:
2x^2 + 2kx + (k^2 - 25) = 0

The discriminant, denoted by Δ, of this quadratic equation is given by Δ = b^2 - 4ac, where a = 2, b = 2k, and c = k^2 - 25.

Now we can analyze the discriminant to determine the number of points of intersection:

1) For two points of intersection: Δ > 0

In this case, we have:
Δ = (2k)^2 - 4(2)(k^2 - 25) > 0
4k^2 - 4(2)(k^2 - 25) > 0
4k^2 - 8k^2 + 200 > 0
-4k^2 + 200 > 0
4k^2 - 200 < 0

Simplifying further:
k^2 - 50 < 0
(k - 5√2)(k + 5√2) < 0

To have two points of intersection, we need the quadratic to be less than zero. Therefore, the values of k that satisfy the condition are -5√2 < k < 5√2.

2) For one point of intersection: Δ = 0

In this case, we have:
Δ = (2k)^2 - 4(2)(k^2 - 25) = 0
4k^2 - 8k^2 + 200 = 0
-4k^2 + 200 = 0
4k^2 = 200
k^2 = 50
k = ±√50 = ±5√2

To have one point of intersection, the discriminant must be zero. Therefore, the values of k that satisfy this condition are k = 5√2 and k = -5√2.

3) For zero points of intersection: Δ < 0

In this case, we have:
Δ = (2k)^2 - 4(2)(k^2 - 25) < 0
4k^2 - 8k^2 + 200 < 0
-4k^2 + 200 < 0
4k^2 > 200
k^2 > 50
k > √50 = 5√2 or k < -√50 = -5√2

Therefore, if k > 5√2 or k < -5√2, there will be no points of intersection between the line y = x + k and the circle x^2 + y^2 = 25.

In summary:
- Two points of intersection: -5√2 < k < 5√2
- One point of intersection: k = 5√2 or k = -5√2
- Zero points of intersection: k > 5√2 or k < -5√2

To find the values of k for which the line y = x + k intersects the circle x^2 + y^2 = 25 at 2 points, 1 point, or 0 points, we need to substitute the equation of the line into the equation of the circle and analyze the resulting quadratic equation.

Let's start with finding the points of intersection for 2 points:

1. Substitute y = x + k into x^2 + y^2 = 25:
x^2 + (x + k)^2 = 25
Simplify this quadratic equation:
x^2 + (x^2 + 2kx + k^2) = 25
Combine like terms:
2x^2 + 2kx + k^2 = 25
Rearrange this equation to the standard quadratic form:
2x^2 + 2kx + (k^2 - 25) = 0

2. The quadratic equation above will have two points of intersection when its discriminant (b^2 - 4ac) is greater than 0:
D = (2k)^2 - 4(2)(k^2 - 25)
D = 4k^2 - 8k^2 + 200
D = -4k^2 + 200

We need D > 0:
-4k^2 + 200 > 0
4k^2 < 200
k^2 < 50
-√50 < k < √50

Therefore, the line y = x + k intersects the circle x^2 + y^2 = 25 at two points for -√50 < k < √50.

Next, let's find the points of intersection for 1 point:

1. The quadratic equation above will have one point of intersection when its discriminant (D) is equal to 0.
D = 0
-4k^2 + 200 = 0
4k^2 = 200
k^2 = 50
k = ±√50

Therefore, the line y = x + k intersects the circle x^2 + y^2 = 25 at one point for k = ±√50.

Finally, let's find the points of intersection for 0 points:

1. The quadratic equation above will have no points of intersection when its discriminant (D) is less than 0.
D < 0
-4k^2 + 200 < 0
4k^2 > 200
k^2 > 50
k > √50 or k < -√50

Therefore, the line y = x + k does not intersect the circle x^2 + y^2 = 25 for k > √50 or k < -√50.

In summary, the line y = x + k intersects the circle x^2 + y^2 = 25 at:
- Two points for -√50 < k < √50
- One point for k = ±√50
- No points for k > √50 or k < -√50.